8 Apr 2023 Shift 1

Cylindrical wire ${\text{m}}=(0.4 \pm 0.01) {\text{g}}$ $ℓ=(8 \pm 0.04) {\text{cm}}$ ${\text{r}}=(6 \pm 0.03) {\text{mm}}$Density $\rho =\;{ {\text{m}}}/{\pi r^2 ℓ} \Rightarrow \rho r^2 ℓ m^{-1}=\;\frac{1}{\pi}=$ const.Differentiating after taking $\log$ on both side$\;\;{ {\text{d}} \rho }/{\rho }+\;{2 {\text{dr}}}/{ {\text{r}}}+\;{ {\text{d}} ℓ}/{ℓ}-\;{ {\text{dm}}}/{ {\text{m}}}=0$ $\;\;\frac{\Delta \rho }{\rho }=\;{\Delta {\text{m}}}/{ {\text{m}}}-\;\frac{\Delta ℓ}{ℓ}-\;{2 \Delta {\text{r}}}/{ {\text{r}}}$ $\; (\;\frac{\Delta \rho }{\rho }) _{\max }= [\;\frac{0.01}{0.4}+\;\frac{0.04}{8}+2 (\;\frac{0.03}{6}) ]$ $\; (\;\frac{\Delta \rho }{\rho }) _{\max }=0.04$ $\;\;\text{ Percentage error }\;=0.04 \times 100=4 \%$Percentage error $=0.04 \times 100=4 \%$

The passenger inside the train is at rest wrt train so frequency heard by passenger inside the train is same as the source frequency i.e., $400 {\text{Hz}}$.

Weight on the earth surface $=m g$ ${\text{mg}}=400 {\text{N}}$ (given)Weight at a depth ${\text{d}} {\text{w}}= {\text{m}} (\;{ {\text{GM}}( {\text{R}}- {\text{d}})}/{ {\text{R}}^3})$ $\; {\text{W}}= {\text{mg}} (1-\;{ {\text{d}}}/{ {\text{R}}})$ $\; {\text{d}}=\;{ {\text{R}}}/{2} \Rightarrow {\text{w}}= {\text{mg}} (1-\;\frac{1}{2}) \Rightarrow {\text{w}}=\;{ {\text{mg}}}/{2}$ $\; {\text{w}}=200 {\text{N}}$

$\;\;\text{ Max. distance covered }\; {\text{d}}=\sqrt{2 {\text{Rh}}_{ {\text{T}}}}$ $\; ( {\text{R}}=\;\text{ radius of earth, }\; {\text{h}}_{ {\text{T}}}=\;\text{ height of antenna }\;)$ $\;\;\text{ Area }\; {\text{A}}=\pi {\text{d}}^2$ $\; {\text{A}}=\pi (2 {\text{Rh}}_{ {\text{T}}})$ $\; {\text{A}}=2 \times 3.14 \times 6400 \times 98 \times 10^{-3}$ $\; {\text{A}} \approx 3942 {\text{km}}^2$

To bring the coil at rest quickly

Dimension of $\;\frac{1}{\mu _0 \epsilon_0}$ $\; {\text{C}}=\;{1}/{{\sqrt{\mu _0 \epsilon_0}}} \Rightarrow \;\frac{1}{\mu _0 \epsilon_0}= {\text{c}}^2$ \;{ [\;\frac{1}{\mu _0 \epsilon_0}] = \[ {\text{c}}^2]$ }$$;= [;{ {\text{L}}^2}/{ {\text{T}}^2}]$

$\; {\text{R}}=\;{ {\text{u}}^2 sin \;2 \theta}/{ {\text{g}}}$ $\; \{ {\text{u}}_1=40 {\text{m}} / {\ s}, \theta_1=30^{\circ}, {\text{u}}_2=60 {\text{m}} / {\ s}, \theta_2=60^{\circ}\}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}= (\;{ {\text{u}}_1}/{ {\text{u}}_2}) ^2 \;\frac{ sin \;2 \theta_1}{ sin \;2 \theta_2}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}= (\;\frac{40}{60}) ^2 \times \;\frac{ sin \;60^{\circ}}{ sin \;120^{\circ}} \Rightarrow \;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{4}{9}$

At steady state current will not be in the capacitor branch.$\; {\text{i}}=\;\frac{4}{6+2+8}$ $\; {\text{i}}=\;\frac{1}{4} {\text{A}}$ $\;\Delta {\text{V}}_{ {\text{C}}_1}= {\text{i}}(6+2)$ $\;\Delta {\text{V}}_{ {\text{C}}_2}= {\text{i}}(2+8)$ $\;\;{\Delta {\text{V}}_{ {\text{C}}_1}}/{\Delta {\text{V}}_{ {\text{C}}_2}}=\;\frac{4}{5}$

Path will be helical with axis along uniform ${ {\text{B}}}^{\to }-$.

Debroglie wavelength $\lambda =\;{ {\text{h}}}/{ {\text{p}}}$ $\;\lambda _{ {\text{p}}}=\lambda _{ {\text{e}}}$ $\;\;{ {\text{h}}}/{ {\text{p}}_{ {\text{p}}}}=\;{ {\text{h}}}/{ {\text{p}}_{ {\text{e}}}} \Rightarrow \;{ {\text{p}}_{ {\text{p}}}}/{ {\text{p}}_{ {\text{e}}}}=1$

Electric field due to uniformly charged insulating solid sphere

$\;\text{bo} {E}_{\text{bo} {B}}=\text{bo} {a}_{\text{bo} {v}} \text{bo} {A}-\text{bo} {a}_{\text{bo} {s}} \text{bo} {A}^{2 / 3}-\text{bo} {a}_{\text{bo} {A}} \;{( {\text{A}}-2 {\text{Z}})^2}/{ {\text{A}}^{1 / 3}}- {\text{a}}_{ {\text{c}}} \;{ {\text{Z}}( {\text{Z}}-1)}/{ {\text{A}}^{1 / 3}}+\delta ( {\text{A}}, {\text{Z}})$ Most appropriate is option (4)

{ {\text{V}}}^{\to }= (2 t \hat{i}+3 {\text{t}}^2 { {\text{j}}^{\^}}) {\text{m}} / {\ s}, \;\text{ mass }\; {\text{m}}=500 {\text{gm}} ${\;\text{ Force, }\;}^{\to } { {\text{F}}}^{\to }= {\text{ma}}$ -\;{ {\text{F}}}^{\to }=\;\frac{1}{2} (\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}}) \Rightarrow { {\text{F}}}^{\to }=\;\frac{1}{2}(2 { {\text{i}}^{\^}}+6 { {\text{t}}^{\^}}) \;{ {\text{F}}}^{\to }=({ {\text{i}}^{\^}}+3 { {\text{t}}^{\^}}) \;\;\text{ At }\; t=1 {\ s} \Rightarrow { {\text{F}}}^{\to }=({ {\text{i}}^{\^}}+3 { {\text{j}}^{\^}}) $\; {\text{x}}=3$

For satellite K.E. $=\;\frac{1}{2} {\ mv}^2=\;\frac{1}{2} {\text{m}} (\sqrt{\;{ {\text{GM}}}/{ {\text{r}}}}) ^2$ ${\text{K}} . {\text{E}} .=\;{ {\text{GMm}}}/{2 {\text{r}}}$Potential energy $U=-\;{ {\text{GMm}}}/{ {\text{r}}}$Total energy $= {\text{K}} . {\text{E}}+ {\text{U}}$ $\;E=-\;\frac{G M m}{2 r}$ $\;U=2 E \;\; \;\text{ St I }\;- \;\text{ incorrect }\;$ $\;\;\text{ K.E. }\;=| {\text{E}}| {\text{St}} \;\text{ II - incorrect }\;$

$\;{| { {\text{F}}}^{\to }_1 |}= {\text{A}},{| { {\text{F}}}^{\to }_2 |}=\;{ {\text{A}}}/{2} \;\; \theta=\;\frac{\pi}{2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\sqrt{ {\text{F}}_1^{2+} {\text{F}}_2^2}$ $\;=\sqrt{ {\text{A}}^{2+} (\;{ {\text{A}}}/{2}) ^2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\;{{\sqrt{5}} {\text{A}}}/{2}$

Aluminium rod Young's modulus$\; {\text{y}}=7.0 \times 10^{10} \;{ {\text{N}}}/{ {\text{m}}^2}$ $\;\;\text{ strain }\; 0.04 \%$ $\;\;\text{ strain }\;=\;\frac{0.04}{100}$Energy per unit volume $=\;\frac{1}{2}$ stress $\times$ strain$\;=\;\frac{1}{2} {\text{y}} \;\text{ strain }\; \times \;\text{ strain }\;$ $\;=\;\frac{1}{2} {\text{y}}(\;\text{ strain }\;)^2$ $\;=\;\frac{1}{2} \times 7 \times 10^{10} \times (\;\frac{0.04}{100}) ^2$Energy per unit volume $=5600 \;{ {\text{J}}}/{ {\text{m}}^3}$

St I FalseEx. in isothermal process temp. is constant but heat can be added. ST II True ${\text{w}}=\int {\text{PdV}}$If volume increases the ${\text{w}}=+$ ve

Pressure at surface $= {\text{P}}_{ {\text{atm}}}=1 \times 10^5 {\text{Pa}}$ ${\text{v}}_{\;\text{surface }\;}=$ ?Pressure at ${\text{h}}=40 {\text{m}}$ depth$\; {\text{P}}= {\text{P}}_{ {\text{atm}}}+\rho {\text{gh}}$ $\; {\text{P}}=10^{5+}10^3 \times 10 \times 40$ $\; {\text{P}}=5 \times 10^5 {\text{Pa}}$ $\; {\text{V}}=1 {\text{cm}}^3$Temp. is constant$\; {\text{P}}_1 {\text{V}}_1= {\text{P}}_2 {\text{V}}_2$ $\;10^5 \times v=5 \times 10^5 \times 1$ $\;v=5 {\text{cm}}^3$

To move the eye piece outside the telescopic tube

$\; {\text{K}}_{ {\text{i}}}=\;{ {\text{P}}_{ {\text{i}}}^2}/{2 {\text{m}}}$ $\; {\text{K}}_{ {\text{f}}}=\;{ ( {\text{P}}_{ {\text{i}}}+{ {\text{P}}_{ {\text{i}}}}/{2}) ^2}/{2 {\text{m}}} \Rightarrow {\text{K}}_{ {\text{f}}}=\;\frac{9}{4} \;{ {\text{P}}_{ {\text{i}}}^2}/{2 {\text{m}}}$ $\;\;\text{ Percentage increase \in K.E. }\;=\;{ {\text{K}}_{ {\text{f}}}- {\text{K}}_{ {\text{i}}}}/{ {\text{K}}_{ {\text{i}}}} \times 100$ $\; \;\;\;\; \;\frac{\;\frac{9}{4}-1}{1} \times 100$ $\;=\;\frac{5}{4} \times 100=125 \%$

$\;\;\text{ Gain \in binding energy }\;=B \cdot E_f-B_{ {\text{i}}}$ $\;=2(121 \times 8.1)-242 \times 7.6$ $\;=121 {\text{MeV}}$

$\; {\text{W}}_{\;\text{ext }\;}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}} \;\; \{ {\text{U}}=-{ {\text{P}}}^{\to } \cdot { {\text{E}}}^{\to }\}$ $\;=- {\text{PE}} \cos \pi-(- {\text{PE}} \cos 0)$ $\;=2 {\text{PE}}$ $\;=2 \times 6 \times 10^{-6} \times 1.5 \times 10^3$ $\;=18 {\text{mJ}}$

Open organ pipe $ℓ=40 {\text{cm}}$Speed of sound ${\text{v}}=360 {\text{m}} / {\ s}$Frequency of second harmonics $f_2=\;{2 {\text{v}}}/{2 ℓ}$ $\; {\text{f}}_2=\;{ {\text{v}}}/{ℓ} \Rightarrow {\text{f}}_2=\;\frac{360}{0.4}$ $\; {\text{f}}_2=900 {\text{Hz}}$

$\;I=\int d m R^2 \Rightarrow R^2 \int {\text{dm}}= {\text{MR}}^2$ $\; {\text{I}}= {\text{MR}}^2$ $\;\;\text{ Given }\; {\text{I}}=\;{1}/{ {\text{x}}} M {\text{M}}^2$ $\; {\text{x}}=1$

$\;d=2+16$ $\;d=18 {\text{cm}}$

${\text{H}}=1.6 \times 10^3 {\text{A}} / {\text{m}}, {\text{n}}=8$ per ${\text{cm}}=800$ per ${\text{m}}$ ${\text{H}}= {\text{nI}} \Rightarrow {\text{I}}=\;{ {\text{H}}}/{ {\text{n}}}$ ${\text{I}}=\;\frac{1.6 \times 10^3}{8 \times 10^2} \Rightarrow {\text{I}}=2 {\text{A}}$

$\; {\text{I}}={neAV_{d }}$ $\; {\text{V}}_{ {\text{d}}}=\;{ {\text{I}}}/{ {\text{neA}}} \Rightarrow V_d=\;\frac{2}{2 \times 10^{28} \times 1.6 \times 10^{-19} \times 25 \times 10^{-6}}$ $\; {\text{V}}_{ {\text{d}}}=25 {\text{m}} / {\ s}$

$\;\;\text{ LC oscillation }\; {\text{L}}=75 {\text{mH}}$ $\; {\text{C}}=1.2 \mu {\text{F}}$ $\; {\text{U}}_{{\max} {\text{L}}}= {\text{U}}_{\max {\text{C}}}$ $\;\;\frac{1}{2} {\text{LI}}_{\max }^2=\;\frac{1}{2} \;{ {\text{q}}_{\max }^2}/{ {\text{C}}}$ $\; {\text{I}}_{\max }=\;{ {\text{q}}_{\max }}/{\sqrt{ {\text{LC}}}} \Rightarrow {\text{I}}_{\max }=\;{2.7 \times 10^{-6}}/{\sqrt{75 \times 10^{-3} \times 1.2 \times 10^{-6}}}$ $\; {\text{I}}_{\max }=9 \times 10^{-3} {\text{A}}$ $\; {\text{I}}_{\max }=9 {\text{mA}}$

For uniform velocity net force $=0$ ${\text{B}}=6 \pi η r {\text{r}}$ $\;\rho \;\frac{4}{3} \pi r^3 g=6 \pi η r v$ $\;η=\;{2 \rho r^2 g}/{9 {\text{v}}}$ $\;η=\;\frac{2 \times 1750 \times (3 \times 10^{-3}) ^2 \times 10}{9 \times 0.35 \times 10^{-2}}$ $\;η=10 {\text{Pa}}- {\ s}$

Anode $\to {\text{H}}_2 \to 2 {\text{H}}^{+}+2 {\text{e}}^{-}$Cathode $\to {\text{AgCl}}+ {\text{e}}^{-} \to {\text{Ag}}+ {\text{Cl}}^{-}$

(a) isoleucine : {\text{CH}}_3- {\text{CH}}_2- {\text{C}_{|}_{\text{CH}_3}\text{H}}- {\text{C}_{|}_{\text{NH}_2}\text{H}}- {\text{COOH}} (b) cysteine \;\; : {\text{HS}}- {\text{CH}}_2- {\text{C}_{|}_{\text{NH}_2}\text{H}}- {\text{COOH}} (c) lysine \;\; : {\text{H}}_2 {\text{N}}- ( {\text{CH}}_2) _4- {\text{C}_{|}_{\text{NH}_2}\text{H}}- {\text{COOH}}(d) methionine \;\; : {\text{CH}}_3- {\text{S}}- {\text{CH}}_2- {\text{CH}}_2- {\text{C}_{|}_{\text{NH}_2}\text{H}}- {\text{COOH}} (e) glutamic acid \;\; : {\text{HOOC}}- {\text{CH}}_2- {\text{CH}}_2- {\text{C}_{|}_{\text{NH}_2}\text{H}}- {\text{COOH}}

$\; {\text{K}}_3 [ {\text{Co}}( {\text{CN}})_6]$ $\;+3+ {\text{x}}-6=0$ $\; {\text{x}}=+3$ $\;⇓$ $\; {\text{Co}}^{+3} \to 3 {\text{d}}^6$ ↿⇂ ↿ ↿ ↿ ↿$\because {\text{CN}}^{-}$is SFL so pairing occur so ↿⇂ ↿⇂ ↿⇂ $\;\text{ u-e }\;=0$ $↓$So diamagnetic

Sn due to Amphoteric nature.

u.e = 4 u.e = 5 u.e = 1 u.e = 2 $[ {\text{Fe}}( {\text{CN}})_6] ^{-3} < [ {\text{Mn}}( {\text{CN}})_6] ^{-3} < [ {\text{CoF}}_6] ^{-3} < [ {\text{MnBr}}_4] ^{-2}$

( {\text{CO}}+ {\text{H}}_2) + {\text{H}}_2 \;{{\to }_{ {\text{ZnO}}, {\text{Cr}}_2 {\text{O}}_3}^{700\text{K}}}_{↓}_{\text{Catalyst}} { {\text{CH}}_3 {\text{OH}}

${\text{n}}$ factor of ${\text{IO}}_3{ }^{-}$and ${\text{I}}^{-}$in the given redox reaction are 5 and 1 respectively. Therefore, ${\text{IO}}_3{ }^{-}$will always react in the molar ratio $1: 5$ to get ${\text{I}}_2$.${\text{IO}}_3^{-}+6 {\text{H}}^{+}+5 {\text{I}}^{-} \to 3 {\text{I}}_2+3 {\text{H}}_2 {\text{O}}$To get 6 molar ${\text{I}}_2$, multiple equation by 2$\;2 {\text{IO}}_3^{-}+12 {\text{H}}^{+}+10 {\text{I}}^{-} \to 6 {\text{I}}_2+6 {\text{H}}_2 {\text{O}}$ $\;\;\text{ So, }\; {\text{x}}=10$

${\text{CaSO}}_4 \cdot 2 {\text{H}}_2 {\text{O}}$ GypsumTo slow down the process of setting.Gypsum is added to control the 'setting of cement'. If not added, the cement will set immediately after miximg of water leaving no time the concrete placing.

Note: Lithium borohydride is commonly used for selective reduction of esters and lactones to the corresponding alcohol.

Data based (Maximum allowed ppm ${\text{F}}^{-}$ $< 2 {\text{ppm}}$ ${\text{SO}}_4^{2-}$ $< 5$ ppm ${\text{NO}}_3^{-}$ $< 50$ ppm ${\text{Zn}}$ $< 500 {\text{ppm}}$

At comparable molecular mass, alcohol has higher b.p. than ether due to H-bond, because ${\text{H}}$-bond leads to stronger associated of molecules.

(A) Saccharin $\;\; \to$ First artificial sweetening agent(B) Aspartame $\;\; \to$ Unstable at cooking temperature used in soft drink and cold drink.(C) Alitame $\;\; \to$ High potency sweetener (2000 more sweeter than cane sugar)(D) Sucralose $\;\; \to$ Stable at coocking temperature. Also it does not provide calories.

$\; {\text{C}}(2.5)$ $\; {\text{P}}(2.1)$ $\;\; \Rightarrow {\text{Br}} > {\text{C}} > {\text{At}} > {\text{P}}$ $\;{\text{Br}}(2.8)$ $\;\;\text{ At }\;(2.2)$

Due to small in size ${\text{Li}}$ and ${\text{Mg}}$ do not from superoxide.${\text{Li}}^{+} \ge {\text{Mg}}^{+2}$ - radius$2 {\text{e}}^{-} 10 {\text{e}}^{-}$Due to diagonal relationship.

$\;\frac{x}{m}=K p^{1 / n} \;\; \log \;\frac{x}{m}=\log K+\;\frac{1}{n} \log P$

(Only iodine)$2 {\text{Cu}}^{2+}+4 {\text{I}}^{-} \to {\text{Cu}}_2 {\text{I}}_2+ {\text{I}}_2$

(A)$\to$ Because formed intermediate carbocation formed by ${\text{Br}}_{( {\text{a}})}$ get stabilised by conjugation with phenyl (B) $\to$ Because the intermediate carbocation formed by ${\text{I}}_{( {\text{a}})}$ become more stable by conjugation (C) $\to$ Becouse, we con't remove ${\text{Br}}_{( {\text{a}})}$ from bridge head carbon (Bredt's rule) (D) $\to$ Becouse, formed intermediate by ${\text{Br}}_{( {\text{a}})}, 3^{\circ}$ carbocation is more stable (stability of carbocation $3^{\circ} > 2^{\circ} > 1^{\circ}$ )

$\; {\text{XeF}}_4+ {\text{SbF}}_5 \to [ {\text{XeF}}_3] ^{+} ( {\text{SbF}}_6) ^{-}$ $\; {\text{m}}+ {\text{n}}+ {\text{x}}+ {\text{y}}=3+1+6+1=11$Xenon fluoride act as ${\text{F}}^{-}$donor and ${\text{F}}^{-}$acceptor.

Fact

(B) ${\text{pk}}_{ {\text{n}}}=-\log (4 \times 10^{-10}) =9.4$Indicator range$\Rightarrow {\text{pk}}_{ {\text{In}}} \pm 1$i.e. 8.4 to 10.4(D) In acidic medium, phenolphthalein is in unionized form and is colourless.

Adiabatic wall {no heat exchange between system and surrounding}$\; {\text{C}}_{ {\text{v}}} \times \Delta {\text{T}}= {\text{P}} \times {\text{t}} / {\ sec}$ $\; {\text{C}}_{ {\text{v}}} \times 5=60 \times 100$ $\; {\text{C}}_{ {\text{v}}}=1200$